Will it slide?

I drive a van. When I accelerate the load inside my van of mass M must be accelerated at acceleration A which is the same as the acceleration of my van by a force F1, where F1=MA.

The mass M has a reactive force Mg on the floor of the van and so the force, F2 needed to make it slide is over the floor is μMg where μ is the coefficient of static friction.

My load will slide when F1 is greater than F2. So the load starts to slide when F1=F2 or MA=μMg. In other words after the M's cancel the acceleration at which my load starts to slide is independent of it's mass. All loads will slide when the van has an acceleration greater than μg.

But I know that light things slide all over the place whereas 2 tonne masses of iron tend to stay put.

even though we model it as constant, the coefficients of static friction (not sliding) or dynamic friction (sliding) are not truly constant when the normal force (from weight) exerted between surfaces gets very large. (probably i should say when the pressure, the normal force per unit area, gets very large.) this is because little imperfections in the contact surfaces that dig into the other surface a little.

try the same experiment with a nice smooth and dry surface on the bed of your van like a smooth sheet of plywood and with identical smooth bottom surfaces of the different sized objects. the smoothest and most consistent surface texture the better. i think in this case, you might have less difference in sliding between different sized masses, but be careful with a 2 ton mass.

I agree, but also the heavier the load the greater the force needed to accelerate it. If my original maths is correct then the masses cancel and the acceleration at which the load slides is independent of its mass.

Similar problem with F1 (or Indy etc.) cars. How can they accelerate from a standing start (no wing down force) at an acceleration greater than μg? I'm pretty sure they do, so I'm wrong somewhere!

It is independant of it's mass. The lighter objects must not have the same static coefficient of friction as the iron. Here's a better test: Take two identical plastic containers. leave one empty and put a few bricks in the other one. They will both start to slide around at the same acceleration.

I agree, but also the heavier the load the greater the force needed to accelerate it. If my original maths is correct then the masses cancel and the acceleration at which the load slides is independent of its mass.

Your math is correct (if messy ). Let's revisit:

[tex]a=\frac{F}{m}[/tex]

and also:

[tex]f_{friction}=\mu N[/tex]

So for a constant acceleration, doubling the mass doubles the force but also doubles the friction.

The problem lies in the difficulties of "friction" problems. There are so many variables to consider. For example, in practice, a heavier load probably makes better contact with the bed of your van, so [tex]\mu[/tex] is probably higher than for, say, a light cardboard box whose base is not making contact with the bed of your van in some places. Also, the math does not consider rotational effects on you load. When you take a corner, your load wants to tip over. My math is not sufficient to calculate the effect on the friction formula, but it may well paint a very different picture. The area of the base of different loads is another factor. The weight of a heavy load makes your van ride smoother, therefore the Normal downward force does not change as much due to the acceleration issues of "bouncing" over a surface; remember that it takes more force (usually) to overcome friction from rest than it does to keep that mass moving, so lighter load is more likely to experience a "moment" when static friction is overcome. And let's not forget that the material is a determing factor for [tex]\mu[/tex].

Similar problem with F1 (or Indy etc.) cars. How can they accelerate from a standing start (no wing down force) at an acceleration greater than μg? I'm pretty sure they do, so I'm wrong somewhere!

Once again, the math is not necessarily going to give you a practical answer, since we can't take deal with all the variables. That said...

Work the problem from the "right" direction. What value of [tex]\mu[/tex] gives me enough a to satisfy your condition?

I did a quick search and came up with a weight of 1550Lbs = 703Kg for an Indy car. Good enough. Take [tex]a=9.8m/s^2[/tex]. On a flat, horizontal surface then:

[tex]\mbox{From }F=ma[/tex]

[tex]F=(703)(9.8)=6889.4 [/tex]

That is both the Normal force due to weight contributing to friction, and the horizontal force we need to exceed to satisfy your scenario. Turns out the Indy car can accelerate at better than [tex]9.8m/s^2[/tex] when [tex]\mu>1[/tex].

I had a look at some coefficient of friction tables, and discovered materials like aluminum on aluminum have values above 1. And since we've shown that changing mass has no effect on the acceleration we can achieve before the tires spin, making the Indy car lighter (or heavier) doesn't help. [tex]\mu>1[/tex] is still required.